Question

我正在尝试做一个学校项目，结果只显示我的MySQL表中的1个数据。这是代码：

<?php
include("connect.php");
$view_users_query= "select * from data_cuti"; 
$run= mysql_query($view_users_query);
while($row= mysql_fetch_array($run)){               
    $name=$row['Name'];  
    $startdate=$row['startdate'];
    $finaldate=$row['finaldate'];  
    $within=$row['within'];
    $reason=$row['reason'];
    $remaining=$row['remaining'];
    $status=$row['status'];
    $id=$row['id'];     
}
?>  
    <tr>            
        <td><?php echo $name;  ?></td>
        <td><?php echo $startdate;  ?></td>
        <td><?php echo $finaldate;  ?></td>   
        <td><?php echo $within;  ?></td>
        <td><?php echo $reason;  ?></td> 
        <td><?php echo $remaining; ?></td>
        <td><?php echo $status;  ?></td>
    </tr>                                         
</table>

任何人都可以帮忙告诉我，我错过了什么？

Answer 1

正如其他评论者所说，你需要在while循环中包含输出块，否则你只能从最后一行获得输出。

试试这个：

<table>
<?php
include("connect.php");
$view_users_query= "select * from data_cuti"; 
$run= mysql_query($view_users_query);
while($row= mysql_fetch_array($run)){               
    $name=$row['Name'];  
    $startdate=$row['startdate'];
    $finaldate=$row['finaldate'];  
    $within=$row['within'];
    $reason=$row['reason'];
    $remaining=$row['remaining'];
    $status=$row['status'];
    $id=$row['id'];     
    ?>  
    <tr>            
        <td><?php echo $name;  ?></td>
        <td><?php echo $startdate;  ?></td>
        <td><?php echo $finaldate;  ?></td>   
        <td><?php echo $within;  ?></td>
        <td><?php echo $reason;  ?></td> 
        <td><?php echo $remaining; ?></td>
        <td><?php echo $status;  ?></td>
    </tr>
<?php } ?>                                       
</table>

您使用的所有中间变量都是多余的，因此这相当于更短：

<table>
<?php
include("connect.php");
$view_users_query= "select * from data_cuti"; 
$run= mysql_query($view_users_query);
while($row= mysql_fetch_array($run)){                   
    ?>  
    <tr>            
        <td><?php echo $row['Name'];  ?></td>
        <td><?php echo $row['startdate'];  ?></td>
        <td><?php echo $row['finaldate'];  ?></td>   
        <td><?php echo $row['within'];  ?></td>
        <td><?php echo $row['reason'];  ?></td> 
        <td><?php echo $row['remaining']; ?></td>
        <td><?php echo $row['status'];  ?></td>
    </tr>
<?php } ?>                                       
</table>

或者更简洁，没有html和php模式之间的混乱切换：

<?php
include("connect.php");
$view_users_query= "select * from data_cuti"; 
$run= mysql_query($view_users_query);
echo "<table>";
while($row= mysql_fetch_array($run)){                    
    echo "<tr>";
    echo "<td>$row['Name']</td>";
    echo "<td>$row['startdate']</td>";
    echo "<td>$row['finaldate']</td> ";  
    echo "<td>$row['within']</td>";
    echo "<td>$row['reason']</td>";
    echo "<td>$row['remaining']</td>";
    echo "<td>$row['status']</td>";
    echo "</tr>";
}                                     
echo "</table>";

Answer 2

如果您也想要标题，可以执行以下操作：

<?php include("connect.php");?>
<table>
<tr>
<th>Name</th>
<th>Start Date</th>
<th>Final Date</th>
<th>Within</th>
<th>Reason</th>
<th>Remaining</th>
<th>Status</th>
</tr>
<?php
$view_users_query= "select * from data_cuti"; 
$run= mysql_query($view_users_query) or die(mysql_error());
?>
<?php 
while($row= mysql_fetch_array($run)){                   
?>  
<tr>            
    <td><?php echo $row['Name'];  ?></td>
    <td><?php echo $row['startdate'];  ?></td>
    <td><?php echo $row['finaldate'];  ?></td>   
    <td><?php echo $row['within'];  ?></td>
    <td><?php echo $row['reason'];  ?></td> 
    <td><?php echo $row['remaining']; ?></td>
    <td><?php echo $row['status'];  ?></td>
  </tr>
<?php } ?>                                       
</table>

需要帮助我的PHP编码

2 个答案: